In what follows I use $aR\,b$ as an abbreviation for $(a,b)\in R$.
To attack problems like these, start with a small relation that satisfies the positive condition. For your second problem, for instance, you could begin by letting $1 R\,2$, $2 R\,3$, and $1R\,3$; if you stop there, you have a transitive relation. Is it reflexive? Is it symmetric? Can you stop at this point?
Of course, you could have started simply with $1R\,1$, which is also transitive, but it’s clearly symmetric, so that’s a bad idea. The same objection rules out starting with $\varnothing$, the empty relation, in which nothing is related to anything: it’s also symmetric. The next simplest way to get transitivity is the one that I actually used.
The same approach works for your first problem and yields the answer suggested by JDH: starting with $1R\,1$ isn’t very helpful, but starting with $1R\,2$ and $2R\,1$ works without further modification.